# Question c0369

Nov 30, 2015

["H"_3"O"^(+)] = ["OH"^(-)] = 2.35 * 10^(-7)"M"

#### Explanation:

The balanced chemical equation for water's self-ionization reaction looks like this

$2 {\text{H"_2"O"_text((l]) rightleftharpoons "H"_3"O"_text((aq])^(+) + "OH}}_{\textrm{\left(a q\right]}}^{-}$

The equilibrium constant for this reaction would look like this

${K}_{e q} = \left({\left[\text{H"_3"O"^(+)] * ["OH"^(-)])/(["H"_2"O}\right]}^{2}\right)$

Now, when dealing with dilute aqueous solutions, the concentration of water can be thought of as being constant. This means that you can write

overbrace(K_(eq) * ["H"_2"O"]^(2))^(color(blue)(K_W)) = ["H"_3"O"^(+)] * ["OH"^(-)]

This expression, ${K}_{e q} \cdot {\left[\text{H"_2"O}\right]}^{2}$, is called water's self-ionization constant, or water's ion product constant.

Now, you know that this constant is equal to $5.5 \cdot {10}^{- 14}$ at ${50}^{\circ} \text{C}$. Since the self-ionization reaction produces equal number of moles of hydronium and hydroxide ions, you can say that $x$ represents the molarity of these species in aqueous solution.

This means that you can write

${K}_{W} = x \cdot x = {x}^{2}$

In your case, you would have

$5.5 \cdot {10}^{- 14} = {x}^{2} \implies x = \sqrt{5.5 \cdot {10}^{- 14}} = 2.35 \cdot {10}^{- 7}$

Therefore,

$\left[\text{H"_3"O"^(+)] = ["OH"^(-)] = color(green)(2.35 * 10^(-7)"M}\right)$

As an interesting comment, the pH of water at this temperature would be

"pH" = - log( ["H"_3"O"^(+)])#

$\text{pH} = - \log \left(2.35 \cdot {10}^{- 7}\right) = 6.63$

The important thing to realize here is that the water would still be neutral because you have equal concentrations of hydronium and hydroxide ions.